How to integrate a function within a square root

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SUMMARY

The discussion focuses on calculating the length of the curve defined by the function x = y^3/30 + 5/(2y) over the interval 3 ≤ y ≤ 5. The integral for the arc length is established as L = ∫ sqrt(1 + (dx/dy)²) dy, leading to the expression sqrt(y^4/100 + 25/(4y^4) + 1/2) dy. The key challenge identified is solving the integral, particularly due to the presence of the square root, with a suggestion to factor out 1/(100y^4) from within the root for simplification.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with derivatives, particularly dx/dy for parametric equations.
  • Knowledge of arc length formulas in calculus.
  • Ability to manipulate algebraic expressions under square roots.
NEXT STEPS
  • Study integration techniques for functions involving square roots.
  • Learn about parametric equations and their derivatives.
  • Explore methods for simplifying expressions under square roots in integrals.
  • Practice solving arc length problems using various functions.
USEFUL FOR

Students studying calculus, particularly those focusing on integration and arc length calculations, as well as educators looking for examples of integrating functions with square roots.

seanm924
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Homework Statement


Find the length of x=y^3/30+5/(2y) on 3 <= y <= 5


Homework Equations



L = integral sqrt(1+(dx/dy)^2)dx

The Attempt at a Solution



I got to the integral upper bound = 5, lower bound = 3, sqrt(y^4/100+25/(4y^4)+1/2)dy. How do I actually solve the integral? It would be find if there wasn't a square root.
 
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Factor out [tex]\frac{1}{100y^4}[/tex] inside the root and see if you can go from there.
 

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